#include"stdio.h"
#include"math.h"
fun(double x,double n)
{
double a,b,c,d,e,g,h,i,j;
double sum=1,0,f;
d=1,a=2,e=2,b=1;
for(;d<=n;d++,a=a*2,e*=2)
{
i=pow(x,a);
j=pow(-1.0,d);
b=i*j;
g=1.0;
for(h=1.0;h<=e;h++)
{
g=g*h;
}
f=b/g;
sum=sum+f;
}
return sum;
}
main()
{
double x,n,sum;
scanf("%f,%f",&x,&n);
sum=fun(x,n);
printf("%f",sum);
}
3 回答
望遠(yuǎn)
TA貢獻1017條經(jīng)驗 獲得超1032個贊
#include"stdio.h"?
#include"math.h"?
double?fun(double?x,double?n)//注意返回值類型
?{?
double?a,b,c,d,e,g,h,i,j;
double?sum=1.0,f;?
d=1,a=2,e=2,b=1;
for(;d<=n;d++,a=a*2,e*=2)
{?
i=pow(x,a);
j=pow(-1.0,d);?
b=i*j;?g=1.0;?
for(h=1.0;h<=e;h++)
{?
g=g*h;
}?
f=b/g;
sum=sum+f;?
}
return?sum;
?}
?main()
?{?
?double?x,n,sum;?
?scanf("%lf,%lf",&x,&n);?//注意double數(shù)據(jù)類型的輸入輸出格式
?sum=fun(x,n);?
?printf("%lf",sum);
?}
Anjaxs
TA貢獻2條經(jīng)驗 獲得超0個贊
#include"stdio.h"?
#include"math.h"?
/**
* (1*x^0)/1!?+?-1*x^2/2!?+?...?+?[-1^(n-1)]*[x^2*(n-1)]/n!??
**/
double?fun(double?x,double?n)?
{?
double?a,?b,?c,?d,?e,?g,?h,?i,?j;?
double?sum?=?1.0,?f;?
d?=?1,?a?=?2,?e?=?2,?b?=?1;?
for(;?d?<=?n;?d++,?a?=?a*2,?e?*=?2)?{
i?=?pow(x,?a);?
j?=?pow(-1.0,?d);?
b?=?i*j;?
g?=?1.0;?
for(h=1.0;?h?<=?e;?h++)?{?
g?=?g*h;?
}?
f?=?b/g;?
sum?=?sum?+?f;?
}?
return?sum;?
}?
main()?
{?
double?x,n,sum;?
scanf("%f,%f",&x,&n);?
sum=fun(x,n);?
printf("%f",sum);?
}可能是這樣
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